Question 868930
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Hi
p(def) = .10,  n = 5
P(X=1) = binompdf(5, .10, 1)= .3281
P (X &#8804; 2)= binomcdf(5, .10,2)= .9914
P(X>3) = 1 - binomcdf(5, .10,3)= .0005